Solving this quadratic using the quadratic formula, x = [-2 ± √(2² + 4*143)] / 2.

["Solving Quadratic Equations Using the Quadratic Formula: A Step-by-Step Guide with Example", "Quadratic equations are fundamental in algebra, appearing frequently in math, physics, engineering, and finance. While some may seem tricky at first glance, solving them using the quadratic formula is straightforward once you break it down. In this article, we’ll explore how to solve quadratics of the form:", "$$\nx = \frac{-2 \pm \sqrt{(2)^2 + 4 \cdot 143}}{2}\n$$", "and explain the importance of each component in the formula.", "---", "### What Is the Quadratic Formula?", "The standard quadratic equation is:", "$$\nax^2 + bx + c = 0\n$$", "The quadratic formula provides the exact solutions for ( x ):", "$$\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n$$", "However, in some variations—especially those involving unique constants—the formula might adapt slightly, such as in this particular example:", "$$\nx = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 143}}{2}\n$$", "This is not the standard discriminant (which is ( b^2 - 4ac )), but an altered expression meant to simplify solving a special case.", "---", "### Step-by-Step Breakdown of the Given Equation", "Let’s rewrite the given expression clearly:", "$$\nx = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 143}}{2}\n$$", "Step 1: Identify Coefficients\nAlthough the formula uses (-2) as the coefficient of (x), compare it to the standard form where the coefficient of (x^2) is (a). Here, from the numerator, we observe:\n- ( b = -2 )\n- The discriminant-like expression is (2^2 + 4 \cdot 143), which aligns with (b^2 + 4ac) only if (4ac = 4 \cdot 143), meaning (a = 143).", "This suggests the equation is likely:", "$$\n143x^2 - 2x + 143 = 0\n$$", "But since the formula shown does not explicitly divide by (2a = 286), we must interpret the expression carefully—possibly the formula is presented in a normalized or specialized variant.", "---", "### Step 2: Compute Key Values", "Start by calculating each part under the square root:", "- ( b^2 = (-2)^2 = 4 )\n- ( 4ac = 4 \cdot 143 \cdot 1 = 572 )? Wait—if (a = 143), then (4ac = 4 \cdot 143 = 572), so:\n [\n \ ext{Discriminant-like term} = b^2 + 4ac = 4 + 572 = 576\n ]", "Check:\n(2^2 + 4 \cdot 143 = 4 + 572 = 576) ✔️", "So,", "$$\nx = \frac{-(-2) \pm \sqrt{576}}{2 \cdot 143} = \frac{2 \pm 24}{286}\n$$", "Note: This confirms the denominator (2a = 2 \cdot 143 = 286), and (\sqrt{576} = 24)", "---", "### Step 3: Solve for Both Roots", "Using the simplified quadratic formula:", "$$\nx = \frac{2 \pm 24}{286}\n$$", "First root:", "$$\nx = \frac{2 + 24}{286} = \frac{26}{286} = \frac{13}{143} = \frac{1}{11}\n$$", "Second root:", "$$\nx = \frac{2 - 24}{286} = \frac{-22}{286} = \frac{-11}{143} = -\frac{1}{13}\n$$", "---", "### Why This Format Appears", "The form shown:", "$$\nx = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 143}}{2}\n$$", "is likely a condensed or reformulated version of the standard quadratic formula, tailored for specific coefficients. It simplifies computation when (b = -2) and (4ac = b^2 + 4ac = 576). While non-standard, recognizing it as a specialized application of the quadratic formula enhances flexibility in solving complex quadratic cases.", "---", "### Final Values", "The two solutions are:", "- ( x = \frac{1}{11} )\n- ( x = -\frac{1}{13} )", "These correspond to the roots of the equation (143x^2 - 2x + 143 = 0), which can be verified via substitution into the original quadratic.", "---", "### Tips for Mastering Quadratic Solutions", "- Always identify (a), (b), and (c) carefully from the standard form (ax^2 + bx + c = 0).\n- The discriminant (D = b^2 - 4ac) tells you about the nature of roots: real, complex, or repeated.\n- Simplify square roots and fractions after computing solutions to express answers clearly.\n- Practice with a range of quadratics—including perfect squares and irrational roots—to build confidence.", "---", "### Conclusion", "Solving quadratics using the quadratic formula is efficient once you recognize coefficient relationships and adapt formulas for specialized cases. The form:", "$$\nx = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 143}}{2}\n$$", "serves as a powerful example of how the quadratic formula can be applied flexibly—even when the discriminant takes a modified form. Whether in academic work or real-world problem-solving, mastering these techniques puts you ahead in algebra and beyond.", "---", "Keywords: quadratic formula, solving quadratics, discriminant, algebra tutorial, quadratic equation solutions, step-by-step quadratic formula, how to use quadratic formula, simplify square roots, math formulas, algebra examples."]